Question

at time $t_1$, an electron is sent along the positive - direction of an x - axis, through both an electric field $vec{e}$ and a magnetic field $vec{b}$, with $vec{e}$ directed parallel to the y - axis. the figure gives the y - component $f_{net,y}$ of the net force on the electron due to the two fields, as a function of the electrons speed v at time $t_1$. the scale of the velocity axis is set by $v_s = 100$ m/s. the x and z components of the net force are zero at $t_1$. assuming $b_x = 0$, find (a) the magnitude of the electric field and (b) $vec{b}$ in unit - vector notation.
(a) number
(b) $hat{i}+hat{j}+hat{k}$
units
units

Explanation:

Step1: Recall the force formula

The net - force on a charged particle in electric and magnetic fields is $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$. Given $F_{x} = F_{z}=0$ and $B_{x} = 0$. The $y$ - component of the force is $F_{net,y}=q(E_{y}+v_{x}B_{z}-v_{z}B_{x})$. Since $B_{x} = 0$, $F_{net,y}=q(E_{y}+v_{x}B_{z})$.

Step2: Find the slope of the $F_{net,y}-v$ graph

The charge of an electron $q=- 1.6\times10^{-19}\ C$. From the graph of $F_{net,y}$ versus $v$, the slope $m$ of the line gives $m = qB_{z}$. The slope of the line passing through $(v = 0,F_{net,y}=-2\times10^{-19}\ N)$ and $(v = 2\ m/s,F_{net,y}=2\times10^{-19}\ N)$ is $m=\frac{\Delta F_{net,y}}{\Delta v}=\frac{(2\times10^{-19}-(-2\times10^{-19}))\ N}{(2 - 0)\ m/s}=2\times10^{-19}\ N/m$.
Since $m = qB_{z}$, then $B_{z}=\frac{m}{q}=\frac{2\times10^{-19}\ N/m}{-1.6\times10^{-19}\ C}=-1.25\ T$. When $v = 0$, $F_{net,y}=qE_{y}$, so $E_{y}=\frac{F_{net,y}}{q}=\frac{-2\times10^{-19}\ N}{-1.6\times10^{-19}\ C}=1.25\ V/m$.

Step3: Calculate the magnitude of the electric field

The magnitude of the electric field $|\vec{E}| = E_{y}=1.25\ V/m$.

Step4: Write the magnetic - field vector

Since $B_{x} = 0$, $B_{y}=0$ (not given any information about $y$ - component of $B$ and from the problem setup) and $B_{z}=-1.25\ T$, $\vec{B}=- 1.25\hat{k}\ T$.

Answer:

(a) $1.25\ V/m$
(b) $\vec{B}=-1.25\hat{k}\ T$